/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 63 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPApplicativeOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) ATransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) UsableRulesProof [EQUIVALENT, 0 ms] (25) QDP (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] (27) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(*, x), app(app(+, y), z)) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z)) APP(app(*, x), app(app(+, y), z)) -> APP(+, app(app(*, x), y)) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, app(app(+, y), z)), x) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z)) APP(app(*, app(app(+, y), z)), x) -> APP(+, app(app(*, x), y)) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y) APP(app(*, app(app(+, y), z)), x) -> APP(*, x) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) APP(app(*, app(app(*, x), y)), z) -> APP(*, y) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, x), app(app(+, y), z)) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z) APP(app(+, app(app(+, x), y)), z) -> APP(+, y) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 17 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, x), app(app(+, y), z)) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, x), app(app(+, y), z)) The TRS R consists of the following rules: app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesReductionPairsProof (EQUIVALENT) First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem. The pairs P are: +1(+(x, y), z) -> +1(y, z) +1(+(x, y), z) -> +1(x, +(y, z)) and the Q and R are: Q restricted rewrite system: The TRS R consists of the following rules: +(+(x, y), z) -> +(x, +(y, z)) Q is empty. By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: +1(+(x, y), z) -> +1(y, z) +1(+(x, y), z) -> +1(x, +(y, z)) The following rules are removed from R: app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(+(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(+1(x_1, x_2)) = 2*x_1 + x_2 ---------------------------------------- (9) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPApplicativeOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04].Here, we combined the reduction pair processor [LPAR04,JAR06] with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem. The a-transformed P is (*1(x, +(y, z)),*1(x, z)) (*1(x, +(y, z)),*1(x, y)) (*1(+(y, z), x),*1(x, y)) (*1(+(y, z), x),*1(x, z)) (*1(*(x, y), z),*1(x, *(y, z))) (*1(*(x, y), z),*1(y, z)) The a-transformed usable rules are *(*(x, y), z) -> *(x, *(y, z)) *(+(y, z), x) -> +(*(x, y), *(x, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) +(+(x, y), z) -> +(x, +(y, z)) The following pairs can be oriented strictly and are deleted. APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z) The remaining pairs can at least be oriented weakly. APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) Used ordering: Polynomial interpretation [POLO]: POL(*(x_1, x_2)) = x_1 + x_1*x_2 + x_2 POL(*1(x_1, x_2)) = x_1 + x_1*x_2 POL(+(x_1, x_2)) = 1 + x_1 + x_2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem. The a-transformed P is *1(x, +(y, z)) -> *1(x, z) *1(x, +(y, z)) -> *1(x, y) *1(*(x, y), z) -> *1(x, *(y, z)) *1(*(x, y), z) -> *1(y, z) The a-transformed usable rules are *(x, +(y, z)) -> +(*(x, y), *(x, z)) *(+(y, z), x) -> +(*(x, y), *(x, z)) *(*(x, y), z) -> *(x, *(y, z)) +(+(x, y), z) -> +(x, +(y, z)) The following pairs can be oriented strictly and are deleted. APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. *1(x1, x2) = x1 *(x1, x2) = *(x1, x2) +(x1, x2) = + Knuth-Bendix order [KBO] with precedence:trivial and weight map: +=1 *_2=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: *(x, +(y, z)) -> *(x, z) *(x, +(y, z)) -> *(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: **(x, +(y, z)) -> *(x, z) The graph contains the following edges 1 >= 1, 2 > 2 **(x, +(y, z)) -> *(x, y) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (22) YES ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The graph contains the following edges 1 >= 1, 2 > 2 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 ---------------------------------------- (27) YES